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zalisa [80]
1 year ago
10

Compare the atomic sizes of sodium and magnesium. Explain. ​

Chemistry
1 answer:
mihalych1998 [28]1 year ago
5 0

<u>Answer:</u>

Sodium atom is larger than Magnesium atom.

<u>Explanation:</u>

Looking at the electronic configurations of Sodium and Magnesium:

_{11}Na: 1s^2 2s^2 p^6 3s^1\\\\_{12}Mg : 1s^2 2s^2 p^6 3s^2 ,

we can see that they both have their valence electrons in the 3rd shell.

But because Magnesium has a higher nuclear charge of +12 because of its 12 protons than that of Sodium's +11, the nucleus of Magnesium attracts its outer electrons more strongly than Sodium does. This makes the Magnesium atom smaller.

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Acetic acid and ethanol react to form ethyl acetate and water, like this:
ladessa [460]

Answer:

1.) Option C is correct.

The rate of reverse reaction is greater than zero, but equal to the rate of the forward reaction.

2) Option B is correct.

The rate of reverse reaction is Greater than zero, but less than the rate of the forward reaction.

3) Option C is correct.

The rate of reverse reaction is Greater than zero, and equal to the rate of the forward reaction.

4) Option A is correct.

How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium? Zero.

Explanation:

HCH,CO2(aq) + C2H5OH(aq) ⇌ C2H,CO2CH3(aq) + H2O

1) Before the main product is removed from the reaction setup, the chemical reaction is at equilibrium.

Chemical equilibrium is a state of dynamic equilibrium such that the concentration of the reactants and the products do not always remain the same but the rate of forward reaction always matches the rate of backward reaction.

2) When 246. mmol of C2HCO2CH3 are removed from the reaction mixture....

And when one of the factors involved in chemical equilibrium changes, Le Chatellier's principle explains that the system then adjusts to remedy this change and takes time to go back to equilibrium again.

When one of the species involved in the chemical reaction at equilibrium, is removed from the reaction mixture, the rate of reaction begins to favour that side of the reaction until equilibrium is re-established.

So, when 246 mmol of one of the products is removed, the response is to cause the rate of forward reaction to be favoured to produce more of products as there are fewer, and the rate of reverse reaction at this moment becomes less than the rate of forward reaction.

3) The rate of the reverse reaction when the system has again reached equilibrium

Like I said in (2) above, the reaction remedies this change in concentration of one of the products until equilibrium is re-established and when chemical equilibrium is re-established the rate of forward reaction once again matches the rate of backward reaction.

4) How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?

By the time equilibrium is re-established, the system goes back to how it all was and the concentration of C2H5CO2CH3 goes back to the same as it was at the start of the reaction.

Hope this Helps!!!

3 0
2 years ago
What are the moles of silver metal produced from 0.0999 mol of copper?
julia-pushkina [17]

Answer:

1234567i9812345678912121212121

6 0
2 years ago
If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
Everyone in squidwards family has light blue skin which is dom trait for body color in hid homtonof squid vally
ArbitrLikvidat [17]
Okay what am i answering?
3 0
2 years ago
If I have 20g of Sodium, how many moles of Sodium Oxide (Na2O) can be made?
Vika [28.1K]
Mass:20g

Mole=20/[2(23)+16]
=0.32 mil
8 0
3 years ago
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